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4.Moving Charges and Magnetism
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An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform of magnetic field $B$. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of
A
$\frac{r}{4}$
B
$\frac{r}{2}$
C
$2r$
D
$4r$
Solution
(d) $r = \frac{{mv}}{{qB}} \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{{{v_1}}}{{{v_2}}} \times \frac{{{B_2}}}{{{B_1}}} \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
${r_2} = 4{r_1}$
Standard 12
Physics
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